What is the solution for y^{''}+a^2y=bcos(cx),y(0)=0,y^'(0)=0 ?

The solution for y^{''}+a^2y=bcos(cx),y(0)=0,y^'(0)=0 is y=(bcos(cx))/(a^2-c^2)-(bcos(ax))/(a^2-c^2)

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y^{''}+a^2y=bcos(cx),y(0)=0,y^'(0)=0

y^{^{''}} + a^{2} y = b cos (c x), y (0) = 0, y^{^{'}} (0) = 0

y = \frac{b cos ( c x )}{a ^{2} - c ^{2}} - \frac{b cos ( a x )}{a ^{2} - c ^{2}}

Solution steps

y^{''} (x) + a^{2} y = b cos (c x)

Solve linear ODE:

y = \frac{b cos ( c x )}{a ^{2} - c ^{2}} - \frac{b cos ( a x )}{a ^{2} - c ^{2}}

y = \frac{b cos ( c x )}{a ^{2} - c ^{2}} - \frac{b cos ( a x )}{a ^{2} - c ^{2}}

What is the solution for y^{''}+a^2y=bcos(cx),y(0)=0,y^'(0)=0 ?
The solution for y^{''}+a^2y=bcos(cx),y(0)=0,y^'(0)=0 is y=(bcos(cx))/(a^2-c^2)-(bcos(ax))/(a^2-c^2)